Question

What are X and Y in the reaction $C_2{H}_4+H_{2}S{O}_4\stackrel{80{}^{o}C}{\to }X(H_{2}O./\Delta a)Y$

• A. $C_2{H}_6,C_2{H}_{5}OH$
• B. $C_2{H}_2,C_2{H}_{5}SH$
• C. $C_2{H}_{5}OS{O}_{3}H,C_2{H}_{5}OH$
• D. $C_2{H}_2,C{H}_{3}CHO$

Answer 1

Answer: (C)

$C_2{H}_4+H_{2}S{O}_4\stackrel{80{}^{o}C}{\to }X\stackrel{H_{2}O\Delta }{\to }Y$ Ethylene reacts with ${\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ to form ethyl hydrogen sulphate as $\underset{C{H}_2}{\stackrel{C{H}_2}{||}}+HHS{O}_4\to \underset{(X)}{\underset{C{H}_{2}HS{O}_4}{\stackrel{C{H}_3}{|}}}$ So, X is ethyl hydrogen sulphate and X (Ethyl hydrogen sulphate) on hydrolysis gives ethyl alcohol and ${\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ is regenerated as - $\underset{C{H}_{2}HS{O}_4+HOH}{\stackrel{C{H}_3}{|}}\stackrel{\Delta }{\to }\underset{\begin{array}{c}\text{Ethyl}\text{alcohol}\\ \text{(}\gamma \text{)}\end{array}}{\underset{C{H}_{2}OH}{\stackrel{C{H}_3}{|}}}+H_{2}S{O}_4$ So, the Y is ethyl alcohol.